证明:先作FE平行AD,且FD平行AC.
∵△ABD与△BEF为相似三角形,∴FB/AB=EB/DB,可变为DB/AB=EB/FB.
又∵AC平行FD,∴DE/CD=FD/AC,可变为DC/AC=DE/FD.
∵DB/AB=EB/FB=DC/AC=DE/FD,∴DB/AB=DC/AC,可变为DB/DC=AB/AC
∴证得DB/DC=AB/AC.
证明:先作FE平行AD,且FD平行AC.
∵△ABD与△BEF为相似三角形,∴FB/AB=EB/DB,可变为DB/AB=EB/FB.
又∵AC平行FD,∴DE/CD=FD/AC,可变为DC/AC=DE/FD.
∵DB/AB=EB/FB=DC/AC=DE/FD,∴DB/AB=DC/AC,可变为DB/DC=AB/AC
∴证得DB/DC=AB/AC.